This thread has been locked.
If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.
Hi Support team
AS D/S page 23, 9.2.2.3 Input Filtering mentioned,
Total effect on gain error can be calculated by replacing the 5-kΩ term with 5 kΩ – 30%, (or 3.5 kΩ) or 5 kΩ +
30%, (or 6.5 kΩ).
my customer just curious why the input impedance will have such higher tolerance about +/-30%?
if so, how about the internal divide resistor, is this part also has +/-30% tolerance?
for example, 3.5Kohm will match with 350kohm & 6.5Kohm will match with 650Kohm? am i correct?
Hi Red,
it's usual for a standard fabrication process that the implemented resistances can show huge production tolerances. But the production tolerances don't have any influence on the ratio of these resistors. Or, by other words, all the resistor values deviate from the nominal values by exactly the same factor. So, the common mode rejection and other important parameters are not influenced by these production tolerances.
Kai
Hi Red,
Kai is exactly correct. In the design and production of resistors inside integrated circuits, it's much easier and more efficient to ensure that the ratio and matching of the internal resistors to one another is tightly controlled, not the absolute value of those resistors. It's the ratio and matching of the resistors that dominates key performance parameters like CMRR, Vos, gain error, etc.
It's for this reason why you should not try to add large resistors to the input network to change closed-loop gain and assume you know what the internal resistances are. It's likely you will not get the gain you want and degrade device performance.
Best regards,
Ian Williams
Applications Manager
Current & Magnetic Sensing